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# A steel pipe has a cross-sectional area of $127.20$ in $^2$ at $25{\degree} \mathrm{F}$. Calculate its crosssectional area when the pipe is heated to $175{\degree} \mathrm{F}$ .

Solution

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$\textbf{Given:}$

$A=127.2 \text{ in}^2$

$T_1= 25 ^\circ \text{F}$

$T_2= 175 ^\circ \text{F}$

$\alpha = 6.5 \times 10^{-6} \text{ } \dfrac{}{^\circ \text{F}}$

The change in surface area is given by the formula:

$\Delta A= 2\cdot \alpha \cdot A \cdot \Delta T$

By including the data given in the task we get:

\begin{align*} \Delta A&= 2 \cdot \alpha \cdot A \cdot (T_2-T_1)\\ &= 2 \cdot 6.5 \times 10^{-6} \cdot 127.2 \cdot (175-25)\\ &\boxed{\Delta A=0.17 \text{ in}^2 } \end{align*}

To calculate the new area we add the initial value and the area change values:

\begin{align*} A_2&=A+\Delta A \\ &=127.2+0.17\\ &\boxed{A_2= 127.37 \text{ in}^2} \end{align*}

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